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yungman
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I am verifying the equation of radiation power of dipole antenna. I found mistakes in the derivation in the notes. I know the final equation is correct. So instead of following the steps in the notes, I reverse the step by using the final formula and going back step by step. Here is the final equation, [tex]P_{rad}=η\frac {|I_0|^2}{4\pi}I_{int}\;\hbox{ where:}[/tex]
[tex]I_{int}=\left[ C +ln(kl)-Ci(kl)+\frac 1 2 \sin(kl)[Si(2kl)-2Si(kl)]+\frac 1 2 \cos(kl)\left(C+ln(\frac{kl}{2})+Ci(2kl)-2Ci(kl)\right)\right][/tex]
Where ##C## is Euler constant and ##C##=0.5772.
The step before this that I want to verify is:
[tex]I_{int}=[1+\cos(kl)]\int_0^{kl}\frac{1-cos(v)}{v}dv-2sin(kl)\int_0^{kl}\frac{\sin(v)}{v}dv+\sin(kl)\int_0^{2kl}\frac{\sin(v)}{v}dv-\cos(kl)\int_0^{2kl}\frac{1-\cos(v)}{v}dv[/tex]
I know
[tex]Si(kl)=\int_0^{kl}\frac{\sin(v)}{v}dv,\;Si(2kl)=\int_0^{2kl}\frac{\sin(v)}{v}dv,\;
Ci(kl)=-\int_0^{kl}\frac{1-\cos(v)}{v}dv+log(kl)+C,\;Ci(2kl)=\int_0^{2kl}\frac{1-\cos(v)}{v}dv+log(2kl)+C[/tex]
I have not idea where [itex]ln(kl), \;ln(\frac{kl}{2})[/itex] come from. How come it is ln instead of log? Also, it will make more sense if it is ##ln(2kl)## rather than ##ln(\frac{kl}{2})##. Please help.
Thanks
[tex]I_{int}=\left[ C +ln(kl)-Ci(kl)+\frac 1 2 \sin(kl)[Si(2kl)-2Si(kl)]+\frac 1 2 \cos(kl)\left(C+ln(\frac{kl}{2})+Ci(2kl)-2Ci(kl)\right)\right][/tex]
Where ##C## is Euler constant and ##C##=0.5772.
The step before this that I want to verify is:
[tex]I_{int}=[1+\cos(kl)]\int_0^{kl}\frac{1-cos(v)}{v}dv-2sin(kl)\int_0^{kl}\frac{\sin(v)}{v}dv+\sin(kl)\int_0^{2kl}\frac{\sin(v)}{v}dv-\cos(kl)\int_0^{2kl}\frac{1-\cos(v)}{v}dv[/tex]
I know
[tex]Si(kl)=\int_0^{kl}\frac{\sin(v)}{v}dv,\;Si(2kl)=\int_0^{2kl}\frac{\sin(v)}{v}dv,\;
Ci(kl)=-\int_0^{kl}\frac{1-\cos(v)}{v}dv+log(kl)+C,\;Ci(2kl)=\int_0^{2kl}\frac{1-\cos(v)}{v}dv+log(2kl)+C[/tex]
I have not idea where [itex]ln(kl), \;ln(\frac{kl}{2})[/itex] come from. How come it is ln instead of log? Also, it will make more sense if it is ##ln(2kl)## rather than ##ln(\frac{kl}{2})##. Please help.
Thanks
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